P2885 [USACO07NOV]电话线Telephone Wire

2021年09月15日 阅读数：3

题目描述

Farmer John's cows are getting restless about their poor telephone service; they want FJ to replace the old telephone wire with new, more efficient wire. The new wiring will utilize N (2 ≤ N ≤ 100,000) already-installed telephone poles, each with some heighti meters (1 ≤ heighti ≤ 100). The new wire will connect the tops of each pair of adjacent poles and will incur a penalty cost C × the two poles' height difference for each section of wire where the poles are of different heights (1 ≤ C ≤ 100). The poles, of course, are in a certain sequence and can not be moved.html

Farmer John figures that if he makes some poles taller he can reduce his penalties, though with some other additional cost. He can add an integer X number of meters to a pole at a cost of X2.ios

Help Farmer John determine the cheapest combination of growing pole heights and connecting wire so that the cows can get their new and improved service.less

输入输出格式

• Line 1: Two space-separated integers: N and Ccode

• Lines 2..N+1: Line i+1 contains a single integer: heighti

• Line 1: The minimum total amount of money that it will cost Farmer John to attach the new telephone wire.

输入输出样例

```5 2
2
3
5
1
4```

`一开始本身写了个DP，竟然能过样例特别兴奋，可是交上去只有70分发现时间复杂度有点高思路比较简单：咱们能够很容易的看出这道题具备无后效性，用dp[i][j]表示前i棵树，第i棵树高度为j的最小代价先预处理一下dp[1][j]，而后对于每一棵树，咱们枚举它的前一棵树的高度和这棵树的高度，计算一下就好时间复杂度n*h*h`
``` 1 #include<iostream>
2 #include<cstdio>
3 #include<cstring>
4 #include<cmath>
5 #include<cstring>
6 #include<algorithm>
7 #include<queue>
8 #include<cstdlib>
9 using namespace std;
10 const int MAXN=100001;
11 const int INF =0x7f7f7f7f;
13 {
14     char c='+';bool flag=0;n=0;
15     while(c<'0'||c>'9'){c=getchar();if(c=='-')flag=1;}
16     while(c>='0'&&c<='9')n=n*10+c-48,c=getchar();
17 }
18 int dp[MAXN][201];// 第i棵树，高度为j的最小花费
19 int n,C;
20 int a[MAXN];
21 int maxheight;
22 int main()
23 {
25     memset(dp,INF,sizeof(dp));
26     for(int i=1;i<=n;i++)
28     for(int i=a[1];i<=maxheight;i++)
29         dp[1][i]=(i-a[1])*(i-a[1]);
30
31     for(int i=2;i<=n;i++)//枚举全部树
32         for(int j=a[i];j<=maxheight;j++)//枚举这棵树的高度
33             for(int k=a[i-1];k<=maxheight;k++)//枚举前一棵树的高度
34                 dp[i][j]=min(dp[i][j],
35                             ((j-a[i])*(j-a[i])+(dp[i-1][k])+abs(j-k)*C));
36
37
38     int ans=0x7fffff;
39     for(int i=a[n];i<=maxheight;i++)
40         ans=min(ans,dp[n][i]);
41     printf("%d",ans);
42     return 0;
43 }```
View Code
`而后化简一下式子`

``` 1 #include<iostream>
2 #include<cstdio>
3 #include<cstdlib>
4 #include<cstring>
5 using namespace std;
6 const int MAXN=300005;
7 const int INF =0x7fffff;
8 const int maxheight=100;
9 int dp[301];// 第i棵树，高度为j的最小花费
10 int f[301];
11 int n,C;
12 int a[MAXN];
13 int bgsum[MAXN];
14 int edsum[MAXN];
15 int main() {
16     scanf("%d%d",&n,&C);
17     for(int i=0; i<n; i++)
18         scanf("%d",&a[i]);
19     memset(dp, 0x3f, sizeof(dp));
20     for(int i=a[0]; i<=maxheight; i++)
21         dp[i]=(i-a[0])*(i-a[0]);
22
23     for(int i=1; i<n; i++) { //枚举全部树
24         memcpy(f,dp,sizeof(dp));
25         for(int j=0; j<=maxheight; j++)    dp[j]=bgsum[j]=edsum[j]=INF;
26
27         bgsum[0]=f[0];
28         for(int j=1; j<=maxheight; j++)
29             bgsum[j]=min(bgsum[j-1],f[j]-C*j);
30
31         edsum[maxheight]=f[maxheight]+maxheight*C;
32         for(int j=maxheight-1; j>=0; j--)
33             edsum[j]=min(edsum[j+1],f[j]+C*j);
34
35         for(int j=a[i]; j<=maxheight; j++) //枚举这棵树的高度
36             dp[j]=min(edsum[j]-C*j,bgsum[j]+C*j)+(j-a[i])*(j-a[i]);
37     }
38     int ans=0x7fffff;
39     for(int i=a[n-1]; i<=maxheight; i++)
40         ans=min(ans,dp[i]);
41     printf("%d",ans);
42     return 0;
43 }```
View Code

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